One-to-Many Relation Composite with Inverse is Coreflexive
Theorem
Let $\RR \subseteq S \times S$ be a relation which is one-to-many.
Then the composite of $\RR$ with its inverse is a coreflexive relation:
- $\RR^{-1} \circ \RR \subseteq \Delta_X$
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That is, by Relation is Symmetric and Antisymmetric iff Coreflexive, $\RR^{-1} \circ \RR$ is both symmetric and antisymmetric.
Proof
As $\RR$ is one-to-many, it follows from Inverse of Many-to-One Relation is One-to-Many that $\RR^{-1}$ is many-to-one.
Let $\tuple {x, z} \in \RR^{-1} \circ \RR$.
Then:
- $\exists y \in S: \tuple {x, y} \in \RR, \tuple {y, z} \in \RR^{-1}$
As $\tuple {x, y} \in \RR$, from the definition of inverse relation:
- $\tuple {y, x} \in \RR^{-1}$
As $\RR^{-1}$ is many-to-one:
- $\tuple {y, z} = \tuple {y, x}$
Thus if $\tuple {x, z} \in \RR^{-1} \circ \RR$, it follows that:
- $\tuple {x, z} = \tuple {x, x}$
Thus:
- $\RR^{-1} \circ \RR \subseteq \Delta_X$
The result follows by definition of coreflexive relation.
$\blacksquare$